Bolzano's theorem, either a lemma or a corollary of the intermediate value theorem, states pretty simply that
If a function $y=f(x)$ is continuous on a closed interval $[a;b]$, and when $f(a)$ and $f(b)$ have opposite signs, then there exists a $c\in(a;b)$ such that $f(c)=0.$
However, every now and again, one might find the alternative statement $c\in[a;b]$. Even Wikipedia is inconsistent; the main thread of the intermediate value theorem is fine with the open interval while the page on the least upper bound property explicitly gives $[a;b].$
Certainly, if Bolzano's theorem holds for $(a;b)$ then for $[a;b]$ the result follows immediately. But the opposite is not true, hence the sanity check.
It is obvious from a commonsense standpoint that the interval should be open, that is $c\in(a;b)$. The problem is thus rigorously proving my intuition. I do not think graphical thinking ought to do it since the reasoning might be circular.
As a final point, the answer may lie in some axiom of real numbers. Sadly, I am not terribly fluent in real analysis to figure it out on my own.
Which is necessary, $(a;b)$ or $[a;b]$? $\leftarrow$ TL; DR
MathWorld, an Internet encyclopedia sponsored by Wolfram Alpha, gives the interval as closed. See this.
Brouwer's fixed point theorem, whose one-dimensional case implies a special case of the intermediate value theorem, claims the interval is, too, $[a;b]$.
Note that if $f(a)$ and $f(b)$ have opposite signs, then $f(a)\not=0\not=f(b)$ - so saying that $f$ has a zero in $[a, b]$ is equivalent to saying that $f$ has a zero in $(a, b)$.